MEMORIAL UNIVERSITY OF NEWFOUNDLAND

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1 MEMORIAL UNIVERSITY OF NEWFOUNDLAND DEPARTMENT OF MATEMATICS AND STATISTICS Worksheet MAT 000 Fall 203 SOLUTIONS (a) First we find any vertical asymptotes We set ( ) 3 = 0 so = Note that the numerator is non-zero, so = is a vertical asymptote To find any horizontal asymptotes, we evaluate 2 ( + 2) ( ) = 3 = = = Thus y = is a horizontal asymptote, and since f() is a rational function, it must approach the horizontal asymptote both as and as To find any -intercepts, we set f() = 0, so 2 ( + 2) = 0, so = 0 or = 2 Thus the points (0, 0) and ( 2, 0) are the -intercepts This also means that (0, 0) is the y-intercept, which we could alternatively find by evaluating f(0) Now we need to find any critical points Note that f () is undefined only when = (the vertical asymptote) so we need only consider f () = 0, that is, ( + 4) = 0, so = 0 or = 4 We can now construct the sign pattern found in Figure We can see that f() is increasing on the interval ( 4, 0) and decreasing on the intervals (, 4 ), (0, ) and (, ) Furthermore, we have a relative minimum at = 4, which is the point ( 4, ) We have a relative maimum at = 0, which is the point (0, 0) 3 3 f () f() + VA 0 4 f () f() + VA + 2 Figure : Sign patterns for Question (a) Finally, we find the hypercritical points Again, f () is undefined only at the vertical asymptote = Furthermore, f () = 0 when 2( + )( + 2) = 0, that is when = or = 2 We therefore construct the sign pattern found in Figure We conclude that f() is concave upward on the intervals ( 2, ) and (, ) and concave downward on the intervals (, 2) and (, ) The points of inflection occur at = 2 and =, which are the points ( 2, 0) and (, 24) Now we can sketch the graph of f(), as found in Figure 2

2 y y = = Figure 2: The graph for Question (a) (b) To find the vertical asymptotes, we set 2 + = 0 so 2 = This has no solutions, implying that there are no vertical asymptotes For the horizontal asymptotes, we observe that f() is rational so we need only take one it at infinity: f() = 2 + giving y = 2 as the horizontal asymptote Now we seek the -intercepts Setting f() = 0 gives = = = 2( ) 2 = 0 so = ; hence (, 0) is the only -intercept Net we evaluate f(0) = 2, so (0, 2) is the y-intercept Now we consider f () Observe that it never fails to eist since the denominator is always positive Thus we set f () = 0 so = 4( + )( ) = 0 giving = and = as the critical points From the sign pattern given in Figure 3 we can see that f() is increasing on (, ) (, ), while it is decreasing on (, ) ence the point (, 4) is a relative maimum while the point (, 0) is a relative minimum Finally, we consider concavity Again, f () never fails to eist so we set f () = 0, giving = 8( 2 3) = 0 = 2,

3 f () f() f () f() Figure 3: The sign patterns for Question (b) Thus = 0, = 3 and = 3 are the hypercritical points From the sign pattern, we see that f() is concave upward on (, 3) (0, 3), and it is concave downward on ( 3, 0) ( 3, ) ence the points ( 3, 2 + 3), (0, 2) and ( 3, 2 3) are all inflection points y 2 Figure 4: The graph for Question (b) We can now sketch the graph depicted in Figure 4 2 (a) First we determine the critical points of f() We have f () = = 3( 2 3) so f () = 0 when = ± 3 and f () is never undefined ence = ± 3 are the only critical points Net we evaluate f() at the critical points and at the endpoints = 4

4 and = 3: f ( ) 3 = 6 ( ) 3 04, f 3 = f( 4) = 28, f(3) = 0 ence the maimum value of f() on [ 4, 3] is 6 3 and the minimum value is 28 (b) First we identify the critical points Note that f () = (2)( + ) ()(2 + 3) ( + ) 2 = ( + 3)( ) ( + ) 2, which is zero for = and = 3, and fails to eist at = For the interval [0, 4], then, the only critical point is =, for which f() = 2 Checking the endpoints, we have f(0) = 3 and f(4) = 9 = 38 ence the maimum value of f() is 9, and the minimum value is 2 (c) As before, we find the critical points Differentiation gives f () = + 2 sin() which fails to eist nowhere, and is zero for = π and = π, both of which are on 6 6 the interval [ π, π] Note that ( f π ) = π ( cos π ) and f ( π 6 ) = π6 2 cos ( π 6 ) 089 At the endpoints, f( π) = π 2 cos( π) 4 and f(π) = π 2 cos(π) 4 ence the maimum value of f() on [ π, π] is approimately 4, while the minimum value is about 226 (d) We again begin by determining the critical points of f(), observing that f () = sec() tan() Setting f () = 0 gives = 0, and f () never fails to eist on the given interval (though it does fail to eist for many other values of ) ence = 0 is the only critical point We evaluate f() there and at the endpoints, giving f(0) =, f ( π ) = , ( π ) f = 2 3 Thus the maimum value of f() on the given interval is 2 and the minimum value is

5 3 Let be the length of fencing parallel to the river and y be the length of the other side of the rectangle The quantity to be maimised is the area A The primary equation is A = y and the secondary equation is so the reduced primary equation is + 2y = 000 = = 000 2y A(y) = (000 2y)y = 000y 2y 2 This problem is defined on an open interval, so we net compute A (y) = 000 4y and set A (y) = 0, giving y = 20 Note that A (y) = 4 so A (y) < 0 for all y, and in particular for y = 20 ence, by the Second Derivative Test, y = 20 is the absolute maimum When y = 20, from the secondary equation we see that = 000 2(20) = 00 Thus the area is a maimum when the plot of land measures 00 metres by 20 metres 4 Let r be the radius of the cylinder (and thus also of the hemisphere) and h be its height The quantity to be minimised is the cost C Note that the surface area of the cylindrical portion (including the bottom) is πr 2 + 2πrh (the normal surface area of a cylinder, minus the surface area of the circle at the top) while the surface area of the hemisphere is 2πr 2 (half the surface area of a sphere) ence the primary equation is C = 2(πr 2 + 2πrh) + 3(2πr 2 ) = 9πr 2 + 4πrh Since the volume of a cylinder is πr 2 h and the volume of a hemisphere is 2π 3 r3 (again, half the volume of a sphere), the secondary equation is πr 2 h + 2π 3 r3 = = h = 2π 3 r3 πr 2 Thus the reduced primary equation is ( 2π ) C(r) = 9πr πr r3 = 9πr πr 2 r 8π 3 r2 = 9π 3 r2 + 4 r Observe that and setting C (r) = 0 yields C (r) = 38π 3 r 4 r 2, 38π 3 r = 4 r 2 = r 3 = 6 9π = r = 3 6 9π

6 Also, C (r) = 38π r 3 = C ( ) 3 6 > 0, 9π so this value of r is the absolute minimum The minimum value of C, then, is ( ) 3 6 C 9π = 9π 3 ( ) π 3 6 9π 290, which means that the cheapest possible cost of the tube is $290 Let and y be the length and height of the poster; see Figure 6 y Figure : A poster with margins, as discussed in Question 3 The quantity to be minimised is its area, A The primary equation is A = y Note that the length of the printed matter, accounting for the side margins, is 8 while the height is y 2 ence the secondary equation is ( 8)(y 2) = 384 = y = The reduced primary equation is ( ) 384 A() = = = A = 2(2 6 92) ( 8) 2

7 We set A = 0 and get = 24 and = 8, but we can disregard the latter because length must be positive Note that A = 644 ( 8) 3 = A (24) > 0 so by the Second Derivative Test, = 24 is an absolute minimum By the secondary equation, when = 24, y = = 36, 24 8 so the dimensions of the poster with the smallest area are 24 cm 36 cm 6 Let the distance between Toontown and the Roadrunner be r, and the distance between Wile E Coyote and Toontown be c The quantity to be minimised is the distance between the Roadrunner and Wile E Coyote, l, as shown in Figure 6 Note that, because he arrives in Toontown at 3:00pm after travelling at km/hr, at 2:00pm Wile E Coyote must be km west of Toontown Coyote Roadrunner Coyote c l r Roadrunner (a) 2:00pm (b) 3:00pm Figure 6: Wile E Coyote fails to catch the Roadrunner, as in Question 4 The primary equation is l = c 2 + r 2 To find secondary equations, let t be the time (measured in hours) elapsed since 2:00pm Then r = 20t and ence the primary equation becomes c = t l(t) = ( t) 2 + (20t) 2 = 62t 2 40t + 22

8 This is defined on the closed interval [0, ], from when Wile E Coyote launches himself towards Toon Town, to when he arrives Observe that l (t) = If we set this equal to zero, we obtain Note that while at the endpoints, (0t 8) 2 2t 2 8t + 9 0t 8 = 0 = t = 9 2 ( ) 9 l = 2, 2 l(0) = and l() = 20 ence the distance between the Roadrunner and Wile E Coyote will be the smallest after 9 2 hours 7 (a) This is a 0 0 indeterminate form: = 0 6 ln(6) 2 ln(2) = ln(6) ln(2) = ln(3) (b) This is a 0 0 indeterminate form: cos ( ) 0 + sin ( ) 2 = 0 + cos ( = 0 + ) 2 sin ( ) = cos ( ) = = 2 (c) This is a 0 0 indeterminate form: sin(m) 0 sin(n) m cos(m) = 0 n cos(n) = m n = m n (d) This is an indeterminate form: ln( + e 2 ) = +e 2 2e 2 2e 2 = = + e 2 4e 2 = 2e 2 = 2 2

9 (e) This is an indeterminate form: [ln()] 3 2 = 3[ln()] 2 2 3[ln()] 2 = ln() = 4 3 ln() = = 4 3 = 4 2 = 0 (f) This is an indeterminate form: 2 + ln() = ( 2 ) = + ln() 2 + ln() = 2 = 2 = (g) This is an 0 indeterminate form: π 2 cos(3) sec(7) cos(3) = π cos(7) 2 (h) This is an indeterminate form: ( ln() ) 3 sin(3) = π 7 sin(7) = 3( ) 7( ) = = ln() ( ) ln() = ( ) + ln() = + ln() = + ln() + = 2 + ln() = 2

10 (i) This is a 0 0 indeterminate form Let y = sin() tan() so ln(y) = tan() ln(sin()) Then ln(sin()) tan() ln(sin()) = cot() = 0 + = 0 +[ sin() cos()] = 0 [cos()] sin() csc 2 () Thus 0 +(sin())tan() = e 0 = (j) This is an 0 indeterminate form Let y = ( + e ) so ln(y) = ln( + e ) Then ln( + ln( + e ) e ) = = ( + e ) +e e = e = + e e = = + e = + e Thus ( + e ) = e = e (k) This is a indeterminate form Let y = cos(3)) so ln(y) = ln(cos(3)) Then ln(cos(3)) ln(cos(3)) = 0 0 Thus 0 (cos(3)) = e 0 = = 0 [ tan(3)] = 0 [ 3 sin(3)] cos(3) = 0 (l) This is a indeterminate form Let y = ( + ) a b ( so ln(y) = b ln + a ) Then ( ) + a ( ) a 2 ( b ln + a ) b ln = Thus ( + a ) b = e ab = b + a 2 = ab + a = ab